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According to the Datasheet of LM7805 (section Application and Implementation), in some scenarios a protective diode (between IN and OUT pins) is needed to protect the regulator.

Please take a look at the schema below, is the protective diode required? I'm asking because the Datasheet say:

may be required if the input is shorted to ground

But in the schema below, the input pin is not shorted to the ground.

schematic

simulate this circuit – Schematic created using CircuitLab

I also tested this circuit on a breadboard and the LM78L05 does not get damaged. During the test the circuit draws ~15mA (seems that 15mA is the current which flow between OUT pin and GND pin).

I want to make sure that the regulator will not get damaged while I'll reprogram the MCU, because while the programmer is connected, the OUT pin of LM7805 will get 5V (while IN pin is floating).

In my opinion in this scenario the protective diode will not help at all, because the input pin is floating (only 2 input capacitors will be charged).

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    \$\begingroup\$ The ST datasheet says: "... protect the device against input short circuit. In this case the input voltage falls rapidly while the output voltage decrease slowly." So it seems the main problem is the output voltage being higher then the input. Under normal circumstances that only happens if the input is shorted or pulled down somehow, but if you are applying 5V to the output using an external source I'd include the diode. \$\endgroup\$
    – Unimportant
    Commented Jul 29, 2022 at 11:53
  • \$\begingroup\$ The diode on output pin will protect the regulator for sure but it also will drop the output voltage. :( \$\endgroup\$
    – Osoian Marcel
    Commented Jul 29, 2022 at 12:06
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    \$\begingroup\$ It won't. The diode is not to be placed in series with the output, it goes between input and output. \$\endgroup\$
    – Unimportant
    Commented Jul 29, 2022 at 12:18
  • \$\begingroup\$ I am new to electronics, so I may not understand some thinks. But the only path for the current to follow is the GND pin of LM7805, because the IN pin is not connected and the capacitors are not a short circuit to ground. So in this specific case (my circuit) I do not understand how a diode may change something. \$\endgroup\$
    – Osoian Marcel
    Commented Jul 29, 2022 at 12:27
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    \$\begingroup\$ The diode costs nothing, and you have one less thing to worry about by adding one. It's well worth doing whenever the input to the regulator is "exposed to the world" while the output is powered. Pretty much a standard item on my design reviews. Some regulators don't need a diode, but this one definitely does. \$\endgroup\$
    – Kuba hasn't forgotten Monica
    Commented Mar 22, 2023 at 2:30

2 Answers 2

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No, it's not required unless there is something else (a heavy load, in particular) connected to the 12V supply. Or if you intend it to be guaranteed to withstand shorting the input.

Adding the diode is not a big deal cost-wise for small quantity, but a failure or reverse installation would bypass the regulator probably destroying whatever it is connected to, and it is one more part to buy and populate.

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  • \$\begingroup\$ Thanks. Exactly, this is the answer I was waiting for. In my particular circuit the input is not connected to any load (or ground), the only load are those 2 capacitors on input. I definitely agree that a protective diode is needed for additional protection (in my case, to charge input capacitors), and I'll add it in my next PCB version. But I do not understand why when I connect 5V on output, the current flows from output to GND pin? The LM7805 does not have any reverse voltage protection between OUT->GND? Is this ok, or the LM7805 may get damaged over time? \$\endgroup\$
    – Osoian Marcel
    Commented Jul 29, 2022 at 17:35
  • \$\begingroup\$ I use a Microchip PICkit3 to program and debug my PIC projects, and I have found that sometimes the regulator will draw enough current that the PICkit shows overcurrent. This may happen if the regulator actually produces, say, 4.8 VDC, which is still in spec, but draws reverse current when 5 VDC is applied. So I set the programmer to 4.75 volts and it works. The PICkit won't supply enough current to damage a 7805. But it might cause problems for some LDOs or micropower devices. \$\endgroup\$
    – PStechPaul
    Commented Jul 29, 2022 at 21:33
  • \$\begingroup\$ It won’t get damaged if the programmer current is limited ( as it typically is by an on board regulator). It will charge the inout capacitors through the regulator. \$\endgroup\$
    – Spehro Pefhany
    Commented Jul 29, 2022 at 21:35
  • \$\begingroup\$ @PStechPaul yes. Presumably the same would happen if there were 100uF+ on the Vdd line directly. \$\endgroup\$
    – Spehro Pefhany
    Commented Jul 29, 2022 at 21:37
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    \$\begingroup\$ @OsoianMarcel Regulators that are rated for voltage externally applied to the output without input may not be that common. Some even have a shunt transistor to short the output when input voltage is removed (to aid in resetting digital parts). Have you found any? Note that things that are perfectly safe at 3.3V out or 5V out may not be safe at 12V due to Vbe breakdown. Here is an example of a circuit for combining 5V inputs (I'm not going to discuss it). \$\endgroup\$
    – Spehro Pefhany
    Commented Aug 1, 2022 at 20:01
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A protection diode is usually recommended in cases where there is enough capacitance on the output of the regulator to cause a reverse voltage from output to input when the input voltage is removed. The diode limits this voltage to around 0.7V.

schematic

simulate this circuit – Schematic created using CircuitLab

You may not have the input being directly grounded, but if the input filter capacitor is discharged and power is connected to the output, current will flow through the regulator to charge that capacitor, and this may damage the regulator, and it might be a gradual thing such that it won't happen the first time, but damage will accumulate until the device eventually fails.

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  • \$\begingroup\$ Thanks for the explanation. I'll definitely add the protection diode (D1) in my next PCB version. \$\endgroup\$
    – Osoian Marcel
    Commented Jul 29, 2022 at 17:36

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